Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1$

So far I've done this:

LHS $$=\cos^{2}3x-\sin^{2}3x$$

$$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$$

$$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$$

I can tell I'm going in the right direction but how should I proceed further?

EDIT I used the identity $$\cos{2x}=2\cos^{2}x-1$$ to solve it in a simpler way. viz.

LHS $$= 2\cos^{2}3x-1$$

$$=2{(4\cos^{3}x-3\cos{x})}^2-1$$

$$2(16\cos^{6}x+9\cos^{2}x-24\cos^{4}x)-1$$

$$=32\cos^{6}x+18\cos^{2}x-48\cos^{4}x-1$$

Still thank you for the answers!

• Use $\sin^2 x = 1 - \cos^2 x$. Commented Jun 16, 2017 at 7:05
• use $\sin^2x=1-\cos^2x$ Commented Jun 16, 2017 at 7:05
• $\sin^4x=(1-\cos^2x)^2$, etc. Commented Jun 16, 2017 at 7:05
• Use $\sin^2x = 1 - \cos^2x$ Commented Jun 16, 2017 at 7:11
• In the general case, the question is related to the Chebishev Polynomials Commented Jun 16, 2017 at 13:26

Another way:

$$\cos(3\cdot2x)=4\cos^3(2x)-3\cos2x$$

Now use $\cos2x=2\cos^2x-1$

Note that you can use the identity $\cos^2 x+\sin^2 x=1$(or alternatively, $\sin^2 x=1-\cos^2 x$).

Take each side of the equation to the power of $2,3$ to get $$9 \sin^2 x=9-9\cos^2 x$$ $$24\sin^4 x=24-48 \cos x^2+24 \cos^4 x$$ $$16\sin^6 x=16-48 \cos^2 x+48 \cos^4 x-16\cos^6 x$$ Tedious, but it probably will work.

• @Arthur Yes, I miswrote that actually. Commented Jun 16, 2017 at 7:25

Simpler way!

Use the fact $\displaystyle (\cos(x)+i\sin(x))^6 = \cos(6x)+i\sin(6x)$.

Expand, use $\sin^2(x)=1-\cos^2(x)$ to simplify higher powers of sine in the real terms, and then organizing by real terms, we get $\cos(6x)+i\sin(6x)=$

$=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)+32i\sin(x)\cos(x)^5-32i\sin(x)\cos(x)^3+6i\sin(x)\cos(x)-1$

Consider the real part.

This gives you $\cos(6x)=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1$.

If you consider the imaginary part, you can isolate $i\sin(6x)$.

This also gives you $\sin(6x)=2\sin(x)\cos(x)^5-32\sin(x)\cos(x)^3+6\sin(x)\cos(x)$

• @Arthur I think it's okay to point out a quicker way of doing the first part. Commented Jun 16, 2017 at 7:16
• @user49640 Then do that in a comment, not an answer. This question is not asking about how to solve $\cos(6x)$, but how to continue the calculations he's already done. Funny when the answers should've been comments, and the comments contain the answer the OP is looking for. Commented Jun 16, 2017 at 7:18
• I don't understand this. Commented Jun 16, 2017 at 7:18
• @mettledmike This uses De Moivre's formula about complex numbers. Commented Jun 16, 2017 at 7:19
• This is because $e^{ix}=\cos(x)+i\sin(x)$, so that implies that $(e^{ix})^6=e^{6ix}=\cos(6x)+i\sin(6x)=(\cos(x)+i\sin(x))^6$ Commented Jun 16, 2017 at 7:20

Uain Prosthaphaeresis Formulas, $$\cos6x+\cos 2x=2\cos4x\cos2x\iff\cos6x=\cos2x(2\cos4x-1)$$

$\cos4x=2\cos^22x-1$ and $\cos2x=2\cos^2x-1$

Expand $(\cos(x)+i\sin(x))^6=\cos(6x)+i\sin(6x)$ and take raeal part of both sides

• If you do that, you get the exact same problem that the OP has already got, namely a lot of $\sin^2$ terms that needs to be taken care of, so this is not helpful. Commented Jun 16, 2017 at 7:13